 Finite-acceleration correction to a relativistic cruise

Replace instantaneous acceleration and braking by legs of constant proper-acceleration magnitude $a$, retaining the same cruise rapidity $\eta$ and fixed endpoint separation. Each leg takes proper time $c\eta /a$ and displaces $(c^2/a)(\cosh \eta -1)$ in the endpoint rest frame. Subtracting the proper cruise time displaced by both legs gives the displayed correction. A cruise segment exists only if the total endpoint separation exceeds twice this acceleration distance.